Problem 124 - Calculus 3

Question

For the following exercises, calculate the partial derivatives. Let \(z=\ln \left(\frac{x}{y}\right) .\) Find \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\)

Step-by-Step Solution

Answer

\(\frac{\partial z}{\partial x} = \frac{1}{x}\), \(\frac{\partial z}{\partial y} = -\frac{1}{y}\).

Step 1: Identify the Function

We are given the function of two variables, \(z = \ln \left(\frac{x}{y}\right)\). Our task is to find the partial derivatives of \(z\) with respect to \(x\) and \(y\).

Step 2: Differentiate with Respect to x

To find the partial derivative \(\frac{\partial z}{\partial x}\), treat \(y\) as a constant. Start by differentiating the function \(z = \ln \left(\frac{x}{y}\right)\) with respect to \(x\). The derivative of \(\ln(u)\) where \(u = \frac{x}{y}\) with respect to \(x\) is \(\frac{1}{u} \cdot \frac{du}{dx}\). Thus, we have:\[\frac{\partial z}{\partial x} = \frac{1}{\left(\frac{x}{y}\right)} \cdot \frac{1}{y} = \frac{y}{x} \cdot \frac{1}{y} = \frac{1}{x}\]

Step 3: Differentiate with Respect to y

To find the partial derivative \(\frac{\partial z}{\partial y}\), treat \(x\) as a constant. Differentiate \(z = \ln \left(\frac{x}{y}\right)\) with respect to \(y\). Here, \(u = \frac{x}{y}\), so apply the derivative \(\frac{1}{u}\frac{du}{dy}\):\[\frac{\partial z}{\partial y} = \frac{1}{\left(\frac{x}{y}\right)} \cdot \left(-\frac{x}{y^2}\right) = -\frac{y}{x} \cdot \frac{x}{y^2} = -\frac{1}{y}\]

Step 4: Summarize Results

Thus, the partial derivatives of \(z\) are \(\frac{\partial z}{\partial x} = \frac{1}{x}\) and \(\frac{\partial z}{\partial y} = -\frac{1}{y}\).

Key Concepts

Multivariable Calculus

In multivariable calculus, we deal with functions of more than one variable. An example of such a function is given in the exercise, where the function depends on two variables:

• The function given is: \( z = \ln \left(\frac{x}{y}\right) \).

Partial derivatives are a crucial part of multivariable calculus. They allow us to understand how changes in one variable affect the function, while keeping the other variables constant.
Here, we compute partial derivatives with respect to the variables \(x\) and \(y\).
By taking these partial derivatives, we can see the rate of change of \(z\) when only \(x\) changes and when only \(y\) changes.
These insights are foundational for analyzing functions that depend on several variables, which is a typical scenario in many real-world applications.

Logarithmic Differentiation

Logarithmic differentiation is a technique used often in calculus to simplify the differentiation process. It is particularly helpful when dealing with functions that are products or quotients, as is the case here with

• \( z = \ln \left(\frac{x}{y}\right) \).

The property of logarithms that allows us to simplify expressions is \( \ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b) \).
This property is very useful because it turns division inside a logarithm into subtraction outside of it, making the differentiation process more straightforward.
By employing this property when differentiating the function with respect to one variable, the calculations become simpler, allowing us to easily identify the resulting partial derivatives.

Chain Rule

The chain rule is another fundamental principle in calculus, especially when dealing with compositions of functions. It is essential when finding the derivatives of functions that are dependent on other functions.

• Here, our function is \( z = \ln(u) \) where \( u = \frac{x}{y} \).

While finding the partial derivatives, the chain rule helps by allowing us to differentiate the outer function, and then multiply it by the derivative of the inner function.
In this exercise, when differentiating \(z = \ln\left(\frac{x}{y}\right)\) with respect to \(x\) or \(y\), we use the chain rule to separate the differentiation into manageable steps.
First, we differentiate \( \ln(u) \), which gives us \( \frac{1}{u} \), then we multiply by the derivative of \( u \) with respect to the desired variable.
This structured approach simplifies the process and is invaluable in higher-level calculus problems.